EMI & Alternating Currents

Concept Core

From Faraday's law to LCR resonance — the complete EMI and AC framework.

Faraday's Laws of Electromagnetic Induction

First Law: An EMF is induced in a conductor whenever the magnetic flux through it changes.

Second Law: The magnitude of the induced EMF is proportional to the rate of change of flux:

ε = −dΦ/dt = −N dΦ/dt (for N-turn coil) Φ = B·A cosθ (magnetic flux) Motional EMF: ε = Bvl (rod moving at speed v in field B, length l)

Lenz's Law: The induced current opposes the change in flux that caused it (the minus sign in Faraday's law).

Self & Mutual Inductance

Self-inductance L: ε = −L·dI/dt Energy stored in inductor: U = ½LI² Solenoid: L = μ₀n²Al (n = turns/length) Mutual inductance M: ε₂ = −M·dI₁/dt

Inductance L depends only on geometry, not on current. Unit: Henry (H) = V·s/A.

AC Circuits — Impedance

Resistor R: V = IR, in phase with I Inductor L: X_L = ωL, V leads I by 90° Capacitor C: X_C = 1/(ωC), V lags I by 90° Series LCR: Z = √(R² + (X_L − X_C)²) Phase angle: tanφ = (X_L − X_C)/R

Resonance in LCR Circuit

At resonance, X_L = X_C:

ω₀ = 1/√(LC) f₀ = 1/(2π√(LC)) At resonance: Z = R (minimum impedance) Current is maximum: I = V/R Power factor = 1 (cos φ = 1)

Quality factor Q = ω₀L/R = 1/(ω₀CR) — measures sharpness of resonance peak.

RMS Values and Power in AC

V_rms = V₀/√2 I_rms = I₀/√2 Average power: P = V_rms I_rms cosφ = I²_rms R Power factor: cosφ = R/Z

For pure inductor or capacitor: cosφ = 0 → P = 0 (no average power consumption). Only resistors dissipate power in AC circuits.

Transformers

V_s/V_p = N_s/N_p = I_p/I_s For ideal transformer: P_in = P_out (100% efficiency) Step-up: N_s > N_p → V_s > V_p, I_s < I_p Step-down: N_s < N_p → V_s < V_p, I_s > I_p

Formula Vault

All EMI and AC circuit formulas for EAPCET.

Faraday's Law

ε = −N dΦ/dt

N = number of turns

Motional EMF

ε = Bvl

Rod length l, speed v, field B

Self-Inductance

ε = −L dI/dt

Unit: Henry (H)

Energy in Inductor

U = ½LI²

Analogue of ½mv² (KE)

Inductive Reactance

X_L = ωL = 2πfL

Increases with frequency

Capacitive Reactance

X_C = 1/(ωC)

Decreases with frequency

Series LCR Impedance

Z = √(R² + (X_L−X_C)²)

Minimum Z = R at resonance

Resonant Frequency

f₀ = 1/(2π√(LC))

X_L = X_C at resonance

RMS Values

V_rms = V₀/√2; I_rms = I₀/√2

Peak/√2 for sinusoidal signals

Power in AC

P = V_rms I_rms cosφ

cosφ = power factor = R/Z

Transformer Ratio

V_s/V_p = N_s/N_p

I_s/I_p = N_p/N_s

Quality Factor

Q = ω₀L/R

Higher Q = sharper resonance

Worked Examples

5 problems — Faraday, self-inductance, LCR impedance, resonance, and transformer.

EasyFind induced EMF: B=0.5T, l=0.2m, v=10 m/s (rod moving perpendicular to B)▾

A rod of length 0.2 m moves at 10 m/s perpendicular to a magnetic field of 0.5 T. Find the induced EMF.

Motional EMF: ε = Bvl = 0.5 × 10 × 0.2 = 1 V

✓ ε = 1 V

EasyEnergy stored in inductor L=50 mH carrying I=2 A▾

Find energy stored in an inductor L = 50 mH when current I = 2 A flows through it.

U = ½LI² = ½ × 50×10⁻³ × 4 = ½ × 0.2 = 0.1 J

✓ U = 0.1 J

MediumFind impedance of series LCR: R=6Ω, X_L=10Ω, X_C=2Ω▾

A series LCR circuit has R=6Ω, X_L=10Ω, X_C=2Ω. Find impedance and phase angle.

Z = √(R² + (X_L−X_C)²) = √(36 + (10−2)²) = √(36+64) = √100 = 10 Ω

tanφ = (X_L−X_C)/R = 8/6 = 4/3 → φ = tan⁻¹(4/3) ≈ 53.1°

Since X_L > X_C, circuit is inductive — voltage leads current.

✓ Z = 10 Ω, φ ≈ 53.1° (inductive)

EAPCET LevelFind resonant frequency of LCR: L=2mH, C=50μF▾

Find the resonant frequency of a series LCR circuit with L = 2 mH and C = 50 μF.

f₀ = 1/(2π√(LC)) = 1/(2π√(2×10⁻³ × 50×10⁻⁶))

LC = 10⁻⁷ → √(LC) = 10⁻³·⁵ = 3.16×10⁻⁴

f₀ = 1/(2π × 3.16×10⁻⁴) = 1/(1.987×10⁻³) ≈ 503 Hz

✓ Resonant frequency f₀ ≈ 503 Hz

Trap QuestionA purely inductive AC circuit consumes power — True or False?▾

An inductor (L only, no resistance) is connected to an AC source. Does it consume power?

The trap: Current flows, voltage exists, so students think power is consumed.

For pure inductor: V leads I by 90° → phase angle φ = 90°.

Power P = V_rms I_rms cosφ = V_rms I_rms × cos90° = V_rms I_rms × 0 = 0 W

The inductor stores energy in its magnetic field during one half-cycle and returns it in the next. No net energy is consumed. Same for pure capacitor (φ = −90°, P = 0).

✓ False — pure L or pure C: P = 0 (power factor = 0); only R dissipates power in AC circuits

Mistake DNA

4 EMI and AC circuit errors from EAPCET distractor analysis.

⚡

Using Peak Values Instead of RMS in Power Formula

AC power P = V_rms × I_rms × cosφ. Using peak values (V₀, I₀) gives twice the correct answer.

❌ Wrong

P = V₀ I₀ cosφ ✗ (gives double the power) (P = ½V₀I₀cosφ is correct but RMS form is easier)

✓ Correct

V_rms = V₀/√2 ✓ I_rms = I₀/√2 ✓ P = V_rms I_rms cosφ ✓

P = V₀I₀cosφ/2 = V_rms I_rms cosφ (both correct). The RMS form is standard. If peak values are given, either divide by 2 at the end or convert to RMS first.

🔄

Lenz's Law: Induced Current Aids the Change

Lenz's law says the induced current OPPOSES the change — it tries to maintain the original flux, not enhance it.

❌ Wrong

Flux increasing → induced current increases flux further ✗ (this would violate energy conservation)

✓ Correct

Flux increasing → induced current opposes ✓ → creates opposing flux ✓ (Lenz's law: oppose, not aid)

Lenz's law is a consequence of energy conservation. If induced current aided the flux change, it would create a self-amplifying system — free energy. It must oppose.

📡

At Resonance: Impedance = 0 (It Equals R)

At resonance X_L = X_C, so Z = √(R² + 0) = R. Impedance is minimum but equals R, not zero.

❌ Wrong

LCR at resonance: Z = 0 (cancel out) ✗

✓ Correct

Z_resonance = R ✓ (X_L and X_C cancel) ✓ Current is maximum V/R ✓ Not infinite current

At resonance, the inductive and capacitive reactances cancel, leaving only R. Impedance = R (minimum). Current = V/R (maximum). Power factor = 1.

🔁

Transformer: High Voltage Side Has More Current

In a step-up transformer (V_s > V_p): N_s > N_p, so I_s < I_p. Higher voltage → lower current (power conservation).

❌ Wrong

Step-up transformer: V_s > V_p → I_s > I_p also ✗ (violates power conservation)

✓ Correct

P = VI = constant ✓ V_s > V_p → I_s < I_p ✓ High voltage side has lower current ✓

Ideal transformer: V_s × I_s = V_p × I_p (power in = power out). Higher voltage on secondary → smaller current on secondary. This is why power is transmitted at high voltage (lower current = lower I²R losses).

Chapter Intelligence

EMI is the bridge between electrostatics, magnetism, and AC circuits — a heavily interconnected chapter.

EAPCET Weightage (2019–2024)

LCR circuit impedance

Resonance: f₀ and Z_min

Faraday's law and EMF

Transformer turns ratio

Power factor and P in AC

Self/mutual inductance

High-Yield PYQ Patterns

Motional EMF = BvlLCR impedance Z calculationResonant frequency f₀ = 1/(2π√LC)Transformer V_s/V_p = N_s/N_pRMS current from peak currentPower factor cosφ = R/ZEnergy stored in inductor ½LI²

Exam Strategy

LCR impedance: Z = √(R² + (X_L−X_C)²). At resonance X_L = X_C so Z = R minimum.
Resonant frequency: f₀ = 1/(2π√LC). Memorise — it's a direct substitution question every year.
Pure L or pure C: cosφ = 0 → P = 0. Only R dissipates power. This is a conceptual question asked frequently.
Transformer: V ratio = N ratio = inverse of I ratio. Higher voltage = lower current (step-up transformer).
This chapter connects directly to Current Electricity (circuit analysis) and Magnetism (inductance = stored magnetic energy).

Concept Core

Formula Vault

Worked Examples

Mistake DNA

Chapter Intelligence

💡 Suggestions & Feedback

Thank you!